You have found the following ages (in years) of all 6 porcupines at your local zoo: $ 4,\enspace 15,\enspace 14,\enspace 6,\enspace 2,\enspace 10$ What is the average age of the porcupines at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{4 + 15 + 14 + 6 + 2 + 10}{{6}} = {8.5\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $4$ years $-4.5$ years $20.25$ years $^2$ $15$ years $6.5$ years $42.25$ years $^2$ $14$ years $5.5$ years $30.25$ years $^2$ $6$ years $-2.5$ years $6.25$ years $^2$ $2$ years $-6.5$ years $42.25$ years $^2$ $10$ years $1.5$ years $2.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{20.25} + {42.25} + {30.25} + {6.25} + {42.25} + {2.25}} {{6}} $ $ {\sigma^2} = \dfrac{{143.5}}{{6}} = {23.92\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{23.92\text{ years}^2}} = {4.9\text{ years}} $ The average porcupine at the zoo is 8.5 years old. There is a standard deviation of 4.9 years.